Chapter 1 Rotation of an Object About a Fixed Axis
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Rotational Motion (The Dynamics of a Rigid Body)
University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Robert Katz Publications Research Papers in Physics and Astronomy 1-1958 Physics, Chapter 11: Rotational Motion (The Dynamics of a Rigid Body) Henry Semat City College of New York Robert Katz University of Nebraska-Lincoln, [email protected] Follow this and additional works at: https://digitalcommons.unl.edu/physicskatz Part of the Physics Commons Semat, Henry and Katz, Robert, "Physics, Chapter 11: Rotational Motion (The Dynamics of a Rigid Body)" (1958). Robert Katz Publications. 141. https://digitalcommons.unl.edu/physicskatz/141 This Article is brought to you for free and open access by the Research Papers in Physics and Astronomy at DigitalCommons@University of Nebraska - Lincoln. It has been accepted for inclusion in Robert Katz Publications by an authorized administrator of DigitalCommons@University of Nebraska - Lincoln. 11 Rotational Motion (The Dynamics of a Rigid Body) 11-1 Motion about a Fixed Axis The motion of the flywheel of an engine and of a pulley on its axle are examples of an important type of motion of a rigid body, that of the motion of rotation about a fixed axis. Consider the motion of a uniform disk rotat ing about a fixed axis passing through its center of gravity C perpendicular to the face of the disk, as shown in Figure 11-1. The motion of this disk may be de scribed in terms of the motions of each of its individual particles, but a better way to describe the motion is in terms of the angle through which the disk rotates. -
L-9 Conservation of Energy, Friction and Circular Motion Kinetic Energy Potential Energy Conservation of Energy Amusement Pa
L-9 Conservation of Energy, Friction and Circular Motion Kinetic energy • If something moves in • Kinetic energy, potential energy and any way, it has conservation of energy kinetic energy • kinetic energy (KE) • What is friction and what determines how is energy of motion m v big it is? • If I drive my car into a • Friction is what keeps our cars moving tree, the kinetic energy of the car can • What keeps us moving in a circular path? do work on the tree – KE = ½ m v2 • centripetal vs. centrifugal force it can knock it over KE does not depend on which direction the object moves Potential energy conservation of energy • If I raise an object to some height (h) it also has • if something has energy W stored as energy – potential energy it doesn’t loose it GPE = mgh • If I let the object fall it can do work • It may change from one • We call this Gravitational Potential Energy form to another (potential to kinetic and F GPE= m x g x h = m g h back) h • KE + PE = constant mg mg m in kg, g= 10m/s2, h in m, GPE in Joules (J) • example – roller coaster • when we do work in W=mgh PE regained • the higher I lift the object the more potential lifting the object, the as KE energy it gas work is stored as • example: pile driver, spring launcher potential energy. Amusement park physics Up and down the track • the roller coaster is an excellent example of the conversion of energy from one form into another • work must first be done in lifting the cars to the top of the first hill. -
M1=100 Kg Adult, M2=10 Kg Baby. the Seesaw Starts from Rest. Which Direction Will It Rotates?
m1 m2 m1=100 kg adult, m2=10 kg baby. The seesaw starts from rest. Which direction will it rotates? (a) Counter-Clockwise (b) Clockwise ()(c) NttiNo rotation (d) Not enough information Effect of a Constant Net Torque 2.3 A constant non-zero net torque is exerted on a wheel. Which of the following quantities must be changing? 1. angular position 2. angular velocity 3. angular acceleration 4. moment of inertia 5. kinetic energy 6. the mass center location A. 1, 2, 3 B. 4, 5, 6 C. 1,2, 5 D. 1, 2, 3, 4 E. 2, 3, 5 1 Example: second law for rotation PP10601-50: A torque of 32.0 N·m on a certain wheel causes an angular acceleration of 25.0 rad/s2. What is the wheel's rotational inertia? Second Law example: α for an unbalanced bar Bar is massless and originally horizontal Rotation axis at fulcrum point L1 N L2 Î N has zero torque +y Find angular acceleration of bar and the linear m1gmfulcrum 2g acceleration of m1 just after you let go τnet Constraints: Use: τnet = Itotα ⇒ α = Itot 2 2 Using specific numbers: where: Itot = I1 + I2 = m1L1 + m2L2 Let m1 = m2= m L =20 cm, L = 80 cm τnet = ∑ τo,i = + m1gL1 − m2gL2 1 2 θ gL1 − gL2 g(0.2 - 0.8) What happened to sin( ) in moment arm? α = 2 2 = 2 2 L1 + L2 0.2 + 0.8 2 net = − 8.65 rad/s Clockwise torque m gL − m gL a ==+ -α L 1.7 m/s2 α = 1 1 2 2 11 2 2 Accelerates UP m1L1 + m2L2 total I about pivot What if bar is not horizontal? 2 See Saw 3.1. -
Kinematics Study of Motion
Kinematics Study of motion Kinematics is the branch of physics that describes the motion of objects, but it is not interested in its causes. Itziar Izurieta (2018 october) Index: 1. What is motion? ............................................................................................ 1 1.1. Relativity of motion ................................................................................................................................ 1 1.2.Frame of reference: Cartesian coordinate system ....................................................................................................................................................................... 1 1.3. Position and trajectory .......................................................................................................................... 2 1.4.Travelled distance and displacement ....................................................................................................................................................................... 3 2. Quantities of motion: Speed and velocity .............................................. 4 2.1. Average and instantaneous speed ............................................................ 4 2.2. Average and instantaneous velocity ........................................................ 7 3. Uniform linear motion ................................................................................. 9 3.1. Distance-time graph .................................................................................. 10 3.2. Velocity-time -
Chapter 5 the Relativistic Point Particle
Chapter 5 The Relativistic Point Particle To formulate the dynamics of a system we can write either the equations of motion, or alternatively, an action. In the case of the relativistic point par- ticle, it is rather easy to write the equations of motion. But the action is so physical and geometrical that it is worth pursuing in its own right. More importantly, while it is difficult to guess the equations of motion for the rela- tivistic string, the action is a natural generalization of the relativistic particle action that we will study in this chapter. We conclude with a discussion of the charged relativistic particle. 5.1 Action for a relativistic point particle How can we find the action S that governs the dynamics of a free relativis- tic particle? To get started we first think about units. The action is the Lagrangian integrated over time, so the units of action are just the units of the Lagrangian multiplied by the units of time. The Lagrangian has units of energy, so the units of action are L2 ML2 [S]=M T = . (5.1.1) T 2 T Recall that the action Snr for a free non-relativistic particle is given by the time integral of the kinetic energy: 1 dx S = mv2(t) dt , v2 ≡ v · v, v = . (5.1.2) nr 2 dt 105 106 CHAPTER 5. THE RELATIVISTIC POINT PARTICLE The equation of motion following by Hamilton’s principle is dv =0. (5.1.3) dt The free particle moves with constant velocity and that is the end of the story. -
Statics FE Review 032712.Pptx
FE Stacs Review h0p://www.coe.utah.edu/current-undergrad/fee.php Scroll down to: Stacs Review - Slides Torch Ellio [email protected] (801) 587-9016 MCE room 2016 (through 2000B door) Posi'on and Unit Vectors Y If you wanted to express a 100 lb force that was in the direction from A to B in (-2,6,3). A vector form, then you would find eAB and multiply it by 100 lb. rAB O x rAB = 7i - 7j – k (units of length) B .(5,-1,2) e = (7i-7j–k)/sqrt[99] (unitless) AB z Then: F = F eAB F = 100 lb {(7i-7j–k)/sqrt[99]} Another Way to Define a Unit Vectors Y Direction Cosines The θ values are the angles from the F coordinate axes to the vector F. The θY cosines of these θ values are the O θx x θz coefficients of the unit vector in the direction of F. z If cos(θx) = 0.500, cos(θy) = 0.643, and cos(θz) = - 0.580, then: F = F (0.500i + 0.643j – 0.580k) TrigonometrY hYpotenuse opposite right triangle θ adjacent • Sin θ = opposite/hYpotenuse • Cos θ = adjacent/hYpotenuse • Tan θ = opposite/adjacent TrigonometrY Con'nued A c B anY triangle b C a Σangles = a + b + c = 180o Law of sines: (sin a)/A = (sin b)/B = (sin c)/C Law of cosines: C2 = A2 + B2 – 2AB(cos c) Products of Vectors – Dot Product U . V = UVcos(θ) for 0 <= θ <= 180ο ( This is a scalar.) U In Cartesian coordinates: θ . -
Position Management System Online Subject Area
USDA, NATIONAL FINANCE CENTER INSIGHT ENTERPRISE REPORTING Business Intelligence Delivered Insight Quick Reference | Position Management System Online Subject Area What is Position Management System Online (PMSO)? • This Subject Area provides snapshots in time of organization position listings including active (filled and vacant), inactive, and deleted positions. • Position data includes a Master Record, containing basic position data such as grade, pay plan, or occupational series code. • The Master Record is linked to one or more Individual Positions containing organizational structure code, duty station code, and accounting station code data. History • The most recent daily snapshot is available during a given pay period until BEAR runs. • Bi-Weekly snapshots date back to Pay Period 1 of 2014. Data Refresh* Position Management System Online Common Reports Daily • Provides daily results of individual position information, HR Area Report Name Load which changes on a daily basis. Bi-Weekly Organization • Position Daily for current pay and Position Organization with period/ Bi-Weekly • Provides the latest record regardless of previous changes Management PII (PMSO) for historical pay that occur to the data during a given pay period. periods *View the Insight Data Refresh Report to determine the most recent date of refresh Reminder: In all PMSO reports, users should make sure to include: • An Organization filter • PMSO Key elements from the Master Record folder • SSNO element from the Incumbent Employee folder • A time filter from the Snapshot Time folder 1 USDA, NATIONAL FINANCE CENTER INSIGHT ENTERPRISE REPORTING Business Intelligence Delivered Daily Calendar Filters Bi-Weekly Calendar Filters There are three time options when running a bi-weekly There are two ways to pull the most recent daily data in a PMSO report: PMSO report: 1. -
Linear and Angular Velocity Examples
Linear and Angular Velocity Examples Example 1 Determine the angular displacement in radians of 6.5 revolutions. Round to the nearest tenth. Each revolution equals 2 radians. For 6.5 revolutions, the number of radians is 6.5 2 or 13 . 13 radians equals about 40.8 radians. Example 2 Determine the angular velocity if 4.8 revolutions are completed in 4 seconds. Round to the nearest tenth. The angular displacement is 4.8 2 or 9.6 radians. = t 9.6 = = 9.6 , t = 4 4 7.539822369 Use a calculator. The angular velocity is about 7.5 radians per second. Example 3 AMUSEMENT PARK Jack climbs on a horse that is 12 feet from the center of a merry-go-round. 1 The merry-go-round makes 3 rotations per minute. Determine Jack’s angular velocity in radians 4 per second. Round to the nearest hundredth. 1 The merry-go-round makes 3 or 3.25 revolutions per minute. Convert revolutions per minute to radians 4 per second. 3.25 revolutions 1 minute 2 radians 0.3403392041 radian per second 1 minute 60 seconds 1 revolution Jack has an angular velocity of about 0.34 radian per second. Example 4 Determine the linear velocity of a point rotating at an angular velocity of 12 radians per second at a distance of 8 centimeters from the center of the rotating object. Round to the nearest tenth. v = r v = (8)(12 ) r = 8, = 12 v 301.5928947 Use a calculator. The linear velocity is about 301.6 centimeters per second. -
Position, Displacement, Velocity Big Picture
Position, Displacement, Velocity Big Picture I Want to know how/why things move I Need a way to describe motion mathematically: \Kinematics" I Tools of kinematics: calculus (rates) and vectors (directions) I Chapters 2, 4 are all about kinematics I First 1D, then 2D/3D I Main ideas: position, velocity, acceleration Language is crucial Physics uses ordinary words but assigns specific technical meanings! WORD ORDINARY USE PHYSICS USE position where something is where something is velocity speed speed and direction speed speed magnitude of velocity vec- tor displacement being moved difference in position \as the crow flies” from one instant to another Language, continued WORD ORDINARY USE PHYSICS USE total distance displacement or path length traveled path length trav- eled average velocity | displacement divided by time interval average speed total distance di- total distance divided by vided by time inter- time interval val How about some examples? Finer points I \instantaneous" velocity v(t) changes from instant to instant I graphically, it's a point on a v(t) curve or the slope of an x(t) curve I average velocity ~vavg is not a function of time I it's defined for an interval between instants (t1 ! t2, ti ! tf , t0 ! t, etc.) I graphically, it's the \rise over run" between two points on an x(t) curve I in 1D, vx can be called v I it's a component that is positive or negative I vectors don't have signs but their components do What you need to be able to do I Given starting/ending position, time, speed for one or more trip legs, calculate average -
Chapters, in This Chapter We Present Methods Thatare Not Yet Employed by Industrial Robots, Except in an Extremely Simplifiedway
C H A P T E R 11 Force control of manipulators 11.1 INTRODUCTION 11.2 APPLICATION OF INDUSTRIAL ROBOTS TO ASSEMBLY TASKS 11.3 A FRAMEWORK FOR CONTROL IN PARTIALLY CONSTRAINED TASKS 11.4 THE HYBRID POSITION/FORCE CONTROL PROBLEM 11.5 FORCE CONTROL OFA MASS—SPRING SYSTEM 11.6 THE HYBRID POSITION/FORCE CONTROL SCHEME 11.7 CURRENT INDUSTRIAL-ROBOT CONTROL SCHEMES 11.1 INTRODUCTION Positioncontrol is appropriate when a manipulator is followinga trajectory through space, but when any contact is made between the end-effector and the manipulator's environment, mere position control might not suffice. Considera manipulator washing a window with a sponge. The compliance of thesponge might make it possible to regulate the force applied to the window by controlling the position of the end-effector relative to the glass. If the sponge isvery compliant or the position of the glass is known very accurately, this technique could work quite well. If, however, the stiffness of the end-effector, tool, or environment is high, it becomes increasingly difficult to perform operations in which the manipulator presses against a surface. Instead of washing with a sponge, imagine that the manipulator is scraping paint off a glass surface, usinga rigid scraping tool. If there is any uncertainty in the position of the glass surfaceor any error in the position of the manipulator, this task would become impossible. Either the glass would be broken, or the manipulator would wave the scraping toolover the glass with no contact taking place. In both the washing and scraping tasks, it would bemore reasonable not to specify the position of the plane of the glass, but rather to specifya force that is to be maintained normal to the surface. -
Chapter 5 ANGULAR MOMENTUM and ROTATIONS
Chapter 5 ANGULAR MOMENTUM AND ROTATIONS In classical mechanics the total angular momentum L~ of an isolated system about any …xed point is conserved. The existence of a conserved vector L~ associated with such a system is itself a consequence of the fact that the associated Hamiltonian (or Lagrangian) is invariant under rotations, i.e., if the coordinates and momenta of the entire system are rotated “rigidly” about some point, the energy of the system is unchanged and, more importantly, is the same function of the dynamical variables as it was before the rotation. Such a circumstance would not apply, e.g., to a system lying in an externally imposed gravitational …eld pointing in some speci…c direction. Thus, the invariance of an isolated system under rotations ultimately arises from the fact that, in the absence of external …elds of this sort, space is isotropic; it behaves the same way in all directions. Not surprisingly, therefore, in quantum mechanics the individual Cartesian com- ponents Li of the total angular momentum operator L~ of an isolated system are also constants of the motion. The di¤erent components of L~ are not, however, compatible quantum observables. Indeed, as we will see the operators representing the components of angular momentum along di¤erent directions do not generally commute with one an- other. Thus, the vector operator L~ is not, strictly speaking, an observable, since it does not have a complete basis of eigenstates (which would have to be simultaneous eigenstates of all of its non-commuting components). This lack of commutivity often seems, at …rst encounter, as somewhat of a nuisance but, in fact, it intimately re‡ects the underlying structure of the three dimensional space in which we are immersed, and has its source in the fact that rotations in three dimensions about di¤erent axes do not commute with one another. -
Rotational Motion of Electric Machines
Rotational Motion of Electric Machines • An electric machine rotates about a fixed axis, called the shaft, so its rotation is restricted to one angular dimension. • Relative to a given end of the machine’s shaft, the direction of counterclockwise (CCW) rotation is often assumed to be positive. • Therefore, for rotation about a fixed shaft, all the concepts are scalars. 17 Angular Position, Velocity and Acceleration • Angular position – The angle at which an object is oriented, measured from some arbitrary reference point – Unit: rad or deg – Analogy of the linear concept • Angular acceleration =d/dt of distance along a line. – The rate of change in angular • Angular velocity =d/dt velocity with respect to time – The rate of change in angular – Unit: rad/s2 position with respect to time • and >0 if the rotation is CCW – Unit: rad/s or r/min (revolutions • >0 if the absolute angular per minute or rpm for short) velocity is increasing in the CCW – Analogy of the concept of direction or decreasing in the velocity on a straight line. CW direction 18 Moment of Inertia (or Inertia) • Inertia depends on the mass and shape of the object (unit: kgm2) • A complex shape can be broken up into 2 or more of simple shapes Definition Two useful formulas mL2 m J J() RRRR22 12 3 1212 m 22 JRR()12 2 19 Torque and Change in Speed • Torque is equal to the product of the force and the perpendicular distance between the axis of rotation and the point of application of the force. T=Fr (Nm) T=0 T T=Fr • Newton’s Law of Rotation: Describes the relationship between the total torque applied to an object and its resulting angular acceleration.